Proving that $\cos\left(ax-\frac{b\pi}{2}\right)=\sin(ax)$ for all $a$, $b$

Question

Prove that $$\cos\left(ax-\frac{b\pi}{2}\right)=\sin(ax)$$ for all $a$, $b$.

I tried expanding the right side using the angle subtraction formula, but that didn't help much. Any help?

Answer 1

The formula is true for all $a$ and $x$ if and only if $b=(4n+1)$ for some integer $n$. [If $b$ is of this type use the formula $\cos(A+B)=\cos (A) \cos(B)-\sin (A) \sin (B)$].

Answer 2

If $$\cos\left(ax-\dfrac{b\pi}2\right)=\sin(ax)=\cos\left(\dfrac{\pi}2-ax\right)$$

$ax-\dfrac{b\pi}2=2m\pi\pm\left(\dfrac{\pi}2-ax\right)$ where $m$ is any integer

$\iff2ax-b=4m\pm(1-2ax)$

$-\implies2ax-b=4m-(1-2x)\iff b=1-4m$

$+\implies2ax-b=4m+1-2ax\iff4ax-b=4m+1$

which is dependent on $x$ hence can not be held for all $a,b$

Answer 3

If $$\cos\left(ax-\frac{b\pi}{2}\right)=\sin(ax)$$ is true, it means that $$f(x)=\cos\left(ax-\frac{b\pi}{2}\right)-\sin(ax)$$ is $0$.

So, consider small values of $x$ and use series expansion to get $$f(x)=\cos \left(\frac{\pi b}{2}\right)+x \left(a \sin \left(\frac{\pi b}{2}\right)-a\right)-\frac{1}{2} x^2 \left(a^2 \cos \left(\frac{\pi b}{2}\right)\right)+\frac{1}{6} x^3 \left(a^3-a^3 \sin \left(\frac{\pi b}{2}\right)\right)+O\left(x^4\right)$$

This would imply $$\cos \left(\frac{\pi b}{2}\right)=0 \implies b=4n+1$$ and if $b=4n+1$ all terms properly cancel.